Linear and Rotational Kinematics System.

AI Thread Summary
The discussion revolves around solving a problem involving linear and rotational kinematics with a block system and a pulley. The user formulates equations based on Newton's second law for the two blocks and the pulley, aiming to find the acceleration of the falling mass. Initial confusion arises regarding the correct representation of torque and the relationship between tension and acceleration. After several edits and recalculations, the user successfully derives the correct formula for acceleration, ultimately arriving at the correct answer of approximately 4.65 m/s². The collaborative effort highlights the importance of correctly applying physics principles to solve complex problems.
CaptainSFS
Messages
57
Reaction score
0

Homework Statement


showme.pl.gif


A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.53. A string attached to the block passes over a pulley to a block of mass m3 = 4 kg. The pulley is a uniform disk of mass m2 = 0.4 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

With what acceleration does the mass m3 fall?

Hint: For each block, apply Newton's 2nd law. For the pulley, apply the rotational analog of Newton's 2nd law. And remember that since the string does not slip on the pulley, there is a relationship between the acceleration of the blocks and the angular acceleration of the pulley. You will end up with three equations and three unknowns (the answers to a,b,c). EDIT: is pulley -> (1/2)m*r*a = T3-T1?

Homework Equations



Kinematics

The Attempt at a Solution



I believe I created the correct formulas for each mass.

Block 1: T1 - Frictional Force = m*a

Block 3: m3*g - T3 = m*a

Pulley 2: Torque = I*alpha -> Torque = (1/2)m2*r*a

I'm however not sure if these are correct, but more importantly, how I would combine them in order to find the acceleration of the falling mass.

I realize that I need 3 equations and unknowns (T1, T2, and a). I believe my Pulley isn't correct because it also has T. Is my assumption correct? And, How would I change the pulley equation so that it works?

EDIT: I believe now that my pulley should be, (1/2)m*r*a=T3 - T1. Is this correct?

EDIT: Still having trouble. Using my above assumptions I came up with this: (m3*g-m1*g*u)/(m3+m1+.5*m2*r)=a
This unfortunately still doesn't work. Any help would be much appreciated, I'm really trying my hardest to figure this out. :)
 
Last edited:
Physics news on Phys.org
Hi CaptainSFS,

CaptainSFS said:

The Attempt at a Solution



I believe I created the correct formulas for each mass.

Block 1: T1 - Frictional Force = m*a

Block 3: m3*g - T3 = m*a

Pulley 2: Torque = I*alpha -> Torque = (1/2)m2*r*a

I'm however not sure if these are correct, but more importantly, how I would combine them in order to find the acceleration of the falling mass.

I realize that I need 3 equations and unknowns (T1, T2, and a). I believe my Pulley isn't correct because it also has T. Is my assumption correct? And, How would I change the pulley equation so that it works?

EDIT: I believe now that my pulley should be, (1/2)m*r*a=T3 - T1. Is this correct?

I don't believe this is quite right. The right hand side (T3-T1) is not the torque. What would the torque be? (Think about what the units of torque are.)
 
torque has units of Nm. So I need some sort of distance I take it? Would I use the radius as this distance? Would it be Torque=(T3-T1)r ?

EDIT: This also doesn't seem to work. I'm feeling pretty lost at this point. =/
 
Last edited:
CaptainSFS said:
torque has units of Nm. So I need some sort of distance I take it? Would I use the radius as this distance? Would it be Torque=(T3-T1)r ?

Yes, that's right.

EDIT: This also doesn't seem to work. I'm feeling pretty lost at this point. =/

Can you show more of your work? What numbers are you using, and what is your final answer?
 
Alright

Block 1: T1 - Frictional Force = m*a

Block 3: m3*g - T3 = m*a

Pulley 2: (1/2)m2*r*a = (T3 - T1)*r

Then, T1 = m1*a + m1*g*u | T3 = m3*g - m3*a <--I plug these two into my 'Pulley 2'

I then algebraically solve for a and I end up with,

a = (m3*g - m1*g*u) / (0.5*m2 + m1 + m3)

I use m1=2kg m2=0.4kg u=.53 m3 = 4kg g=9.81m/ss || The answer I get is 3.51724 m/ss

EDIT: Okay, So I plugged it in again, and I got 4.6518m/ss, which is the right answer. :P

Thanks for your help! :)
 
Glad to help!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top